Sir, in module P test wether all polynomials with integer coefficients are irreducible in rational field? if not , then when we can stop , the process of choosing prime numbers when the degrees are not macthed or f bar function is reducible over integer mod P
If f(x) = X^4 + x^2 + 1 , it is irreducible over z2 hence it should be irreducible over Z. But actually it is reducible over Z. Because (x^2+1+x) (x^2+1-x)
PID par December 2017 Ka Bucklet A, questions no. 85 Ka solution btaiye na sir please, last option me dikat ho rha h sir. I am writing your answer. Sir please help me.
Look, this is the ring in two variables. The ideal (x^2+1,y) is maximal ideal of R[x,y] so given factor ring is a field and hence it is PID.....the logic is same for option 1. For option 2 , since Z is not a field z[x] is not PID. for option 3, R[x,y] is not PID , since R[X] is PID but can never be a field and hence R[x,y] is not a PID....
![Irreducible polynomial of R [x] - Definition -Euclidean Domain - Lesson 45](http://i.ytimg.com/vi/TVvP9yr4YJ0/mqdefault.jpg)
Sir,Aap ka padhane or samjhane ka jo style hai sir, o UNIQUE hai .
so thank you sir
I tried many times to understand p- test but couldn't but now i understood p-test by you.thanks a lot respected sir..
Awesome explain
Very clear explanation
Nice explain dear sir
Sir kindly provide part 3 of irreducibility of polynomial
3no video on irreducibility upload plzzzzz
Bahat Acha explain karteho sir ...thank you....😄
Thankyou so much sir....
Your teaching style is great ....😊😊😊
Sir bhut achha samjhaye aap🙏
Is topic ka 3 lecture nhi dikh rha
very useful sir
sir ap daily 2 or more video upload kar dia karo sir kyoki sir exam se pahle revision ka time mil jayega
Hello Anurag, I try to upload more videos but time problem dear. Here I have to take regular Classes also..... So bit time problem.....
sir apke padane aur samjhane ka tarika bahut hi achha hai. sir pls upload fermats theorem
really sir very good explanation....
sir, aapko bahut bahut dhanybaad.
you are great sir
Outstanding 👏👌👌👍👍
Lecture 3rd nhi mil raha
Really very very thank you sir please complete NET syllabus sir please please
Sir lecture 3 kaha hai, mil nahi Raha,
Thank you sir.
You are very very great
sir lelture3 Ka video private Hai Plz Do public
bhut acche se samjh aaya sir thank u sir
Sir apne ek standard book follow karne bola .Pls suggest that book
Sir, in module P test wether all polynomials with integer coefficients are irreducible in rational field? if not , then when we can stop , the process of choosing prime numbers when the degrees are not macthed or f bar function is reducible over integer mod P
Sir modulo p test polynomial ki kitne degree ke leye true hai
How one can allow to open private videos
Thank you sir ji ❤
Dear student, thank you for your appreciation. Keep learning and keep supporting us..
Irreducibility over Q doesn't imply irreducibility over Z.. it is not iff condition
If f(x) = X^4 + x^2 + 1 , it is irreducible over z2 hence it should be irreducible over Z. But actually it is reducible over Z. Because (x^2+1+x) (x^2+1-x)
Modulo p test bhi field p hi apply hota h ????
Lecture 3 nhi hai kya irreducibility ka??
Sir
👌👌👌
Why must p be prime ?
Sir ur response would be highly respected...
sir group and ring homomorphism find karne ka detail method explain kar dejiye
Sir modulo p test sirf Q k liye hai irreducibility check krne k liye over z(p)
Sir ,kuch polynomial modp test se irreducible nhi hote h but eisenstein test se wo irreducible hote h ..is it possible??
Sir y p test sirf Z(x) p hi apply krege ?
good video
sir isse related net k question ka solution kijiyega sir
Aapka institute Kha hai sir??
F(x)=x^2 - 9 over Z11 what can we say about this function...
Sir jo over Q me irreducible hai oh Z me hota hai but 2x^2+8 sir ye Q pe hai but Z me nai hai
X^4+x^2+1 to esse irreducible aa rha hai..but ye to reducible hai...sir doubt clear karde please..
PID par December 2017 Ka Bucklet A, questions no. 85 Ka solution btaiye na sir please, last option me dikat ho rha h sir. I am writing your answer. Sir please help me.
Look, this is the ring in two variables. The ideal (x^2+1,y) is maximal ideal of R[x,y] so given factor ring is a field and hence it is PID.....the logic is same for option 1. For option 2 , since Z is not a field z[x] is not PID. for option 3, R[x,y] is not PID , since R[X] is PID but can never be a field and hence R[x,y] is not a PID....
Arvind Singh Yadav ,SR institute for Mathematics Thank you sir
Proof kidhar hain!! 😑😑